Sign the Derivative of $(\frac{1-q^r}{1-q^n})^{1/(n-r)}$ wrt $n$ (where $q \in (0,1)$) For $n, r \in \mathbb{N}$, $n > r$ and $q \in (0,1)$, I'm trying to sign the derivative wrt $n$ of 
\begin{eqnarray}
f(n) = \left(\frac{1-q^r}{1-q^n}\right)^{1/(n-r)}.
\end{eqnarray}
I believe the derivative is
\begin{eqnarray}
f'(n) = \left(\frac{1-q^r}{1-q^n}\right)^{\frac{1}{n-r}} \left(\frac{q^n \log (q)}{\left(1-q^n\right) (n-r)}-\frac{\log \left(\frac{1-q^r}{1-q^n}\right)}{(n-r)^2}\right).
\end{eqnarray}
I am very sure that the derivative has the same sign for all admissible values of $x, r,$  and $n$: it should be positive. However proving it is hard. Clearly, $f'(n) > 0$ iff
\begin{eqnarray}
 (n-r)q^n \log (q) > (1-q^n)[\log (1-q^r) - \log(1-q^n)],
\end{eqnarray}
iff
\begin{eqnarray}
q^n \log (q^{n-r}) + (1-q^n)\log(1-q^n) > (1-q^n)\log (1-q^r).
\end{eqnarray}
This look on the right track until you realise 
\begin{eqnarray}
n > r \iff q^n < q^r \iff 1-q^n > 1-q^r \iff \log(1-q^n) > \log(1-q^r).
\end{eqnarray}
Any ideas what I might be missing?
 A: I strongly believe the following to be a solution, however, I would be grateful if somebody could quickly go through my reasoning to make sure I didn't go wrong.
--
Let $A \equiv 1-q^r$. Then
    \begin{eqnarray*}
  f'(n) &=& \frac{\delta}{\delta n}\left( \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \right) A^\frac{1}{n-r} + \frac{\delta}{\delta n}\left( A^\frac{1}{n-r} \right) \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \\
  &=& \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \left[ \frac{q^n \ln q}{(1-q^n)(n-r)} + \frac{\ln(1-q^n)}{(n-r)^2}\right] A^\frac{1}{n-r} - \frac{A^\frac{1}{n-r} \ln A}{(n-r)^2} \left(\frac{1}{1-q^n}\right)^\frac{1}{n-r} \\
  &\propto& (n-r) \frac{q^n \ln q}{1-q^n} + \ln(1-q^n) - \ln(A) =: Z(n).
 \end{eqnarray*}
    Our strategy will be to show that $Z(n) \geq 0$, whence we will know that $f'(n) \geq 0$. To this end, we firstly note
    \begin{eqnarray*}
  Z''(n) &=& \frac{q^n (n-r) \log ^2(q)}{1-q^n}+\frac{q^{2 n} (n-r) \log ^2(q)}{\left(1-q^n\right)^2} \geq 0.
 \end{eqnarray*}
    Thus, in order to show $Z'(n) \geq 0$ it suffices to show $Z'(r+1) \geq 0$, i.e.
    \begin{eqnarray*}
  Z'(r+1) = \frac{q^{r+1} \ln(q)}{1-q^{r+1}} + \ln(1-q^{r+1}) - \ln(1-q^r) \geq 0.
 \end{eqnarray*}
    Now consider
    \begin{eqnarray*}
  \frac{\delta}{\delta q} Z'(r+1) = \frac{q^{r-1} \left(q (r+1) \left(q^r-1\right) \log (q)-(q-1) r \left(q^{r+1}-1\right)\right)}{\left(q^r-1\right) \left(q^{r+1}-1\right)^2}.
 \end{eqnarray*}
    We claim $\frac{\delta}{\delta q} Z'(r+1) \geq 0$ (for $0 < q < 1$). In order to establish this, we solve
    \begin{eqnarray*}
   q^{r-1} \left(q (r+1) \left(q^r-1\right) \log(q)-(q-1) r \left(q^{r+1}-1\right)\right) &=& 0 \\
   \iff \frac{(q-1)(q^{r+1}-1)}{q^r-1} \frac{r}{r+1} = q \log(q) \vee q = 0.
 \end{eqnarray*}
    Now note
    \begin{eqnarray*}
  \frac{(q-1)(q^{r+1}-1)}{q^r-1} &=& - \frac{(1-q)^2(1+ \dots + q^r)}{q(1-q)(1+\dots+q^{r-1})} \\
  &=& - \frac{(1-q)(1+ \dots + q^r)}{1+\dots+q^{r-1}} \\
  &=& -(1-q) \left(1 + \frac{q^r}{1+\dots+q^{r-1}}\right) \leq -(1-q),
 \end{eqnarray*}
    where the inequality holds on $0 < q < 1$. 
On the same domain, we claim $q \log(q) > -(1-q)$. As the LHS tends to $0$ as $q \to 0$ and the RHS tends to $-1$, this is true for at least one point in the domain. Furthermore, the solution to $q \log(q) = -(1-q)$ is given by $q = \exp(1 + W(-e^{-1}))$, where $W(\cdot)$ denotes the Lambert W function. As $W_0(-e^{-1}) = W_{-1}(-e^{-1}) = -1$ (i.e. the two real branches of the Lambert function agree at $-1/e$), we have a unique solution of $q = 1$. Thus, as both functions are continuous, the LHS must lie strictly above the RHS on the domain. 
So the only solutions to $\frac{\delta}{\delta q} Z'(r+1) = 0$ on $q \in [0,1]$ are the boundary points. Thus, for any intermediate points $\frac{\delta}{\delta q} Z'(r+1)$ must have the same sign. Now consider 
    \begin{eqnarray*}
  \frac{\delta}{\delta q}|_{q=1/2} Z'(r+1) = \frac{2^{1-r} \left(\frac{1}{2} \left(2^{-r-1}-1\right) r-\frac{1}{2} \left(2^{-r}-1\right) (r+1) \log (2)\right)}{\left(2^{-r-1}-1\right)^2 \left(2^{-r}-1\right)}.
 \end{eqnarray*}
    It is clear that this is positive iff
    \begin{eqnarray*}
  \frac{1}{2} \left(2^{-r-1}-1\right) r &\leq& \frac{1}{2} \left(2^{-r}-1\right) (r+1) \log (2) \\
   \left(2^{-r-1}-1\right) r &\leq&  \left(2^{-r}-1\right) (r+1) \log (2) \\
   \frac{2^{-r-1}-1}{2^{-r}-1} &\geq& \frac{r+1}{r} \log(2)\\
   \frac{1/2}{1-(1/2)^r} &\geq& \frac{r+1}{r} \log(2)-1.
 \end{eqnarray*}
    Clearly, the LHS is bounded below by $1/2$ for $r\geq 1$, and the RHS is strictly less than $1/2$ for $r \geq 1$. Thus, this inequality holds for all $r \in \mathbb{N}$. Hence, $\frac{\delta}{\delta q} Z'(r+1) \geq 0$. 
Hence, it suffices to show 
    \begin{eqnarray*}
  \lim_{q \to 0} Z'(r+1) &=& \lim_{q \to 0} \frac{q^{r+1} \ln(q)}{1-q^{r+1}} + \ln(1-q^{r+1}) - \ln(1-q^r) \\
  &=& 0.
 \end{eqnarray*}
    in order to establish $Z'(n) \geq 0$. But $f'(n) \geq 0$ iff $Z'(n) \geq 0$, and hence we are done.
